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Chapter 16: Problem 10

If placed in a steady current of \(1 \mathrm{~m} / \mathrm{s}\), and at a generation efficiency of \(40 \%\), how large would an ocean current rotor be (diameter) to satisfy the \(10,000 \mathrm{~W}\) demand of the average American? Put this scale in the context of some familiar object or space.

Short Answer

Expert verified
Answer: The approximate diameter of the ocean current rotor is 7.87 meters, which is roughly the length of a school bus or around half the width of a basketball court.

Step by step solution

01

Identify given parameters

In this problem, we are given: - Power output demand (P) = 10,000 W - Efficiency (η) = 40% - Steady current speed (v) = 1 m/s
02

Calculate actual power generation

The efficiency of the system is given, so we need to calculate the actual power generation (Pg) of the rotor before losses due to efficiency. We can use the following formula: Pg = P ÷ η Where η needs to be converted to its decimal value, which is 0.40. Pg = 10,000 W ÷ 0.40 Pg = 25,000 W
03

Use power formula to find rotor area

The general formula for the power generated from an ocean current is given by: Pg = 0.5 × ρ × A × v³ Where: - Pg is the power generated (25,000 W) - ρ is the density of seawater (approximately 1025 kg/m³) - A is the rotor area (m²), which we want to find - v is the steady current speed (1 m/s) First, rearrange the formula to solve for A: A = Pg ÷ (0.5 × ρ × v³) Next, plug in the given values and calculate A: A ≈ 25,000 ÷ (0.5 × 1025 × 1³) A ≈ 48.78 m²
04

Calculate rotor diameter

Since the rotor is circular, we can relate the rotor area to the diameter of the rotor using the area of a circle formula: A = π × (D/2)² Where: - A is the rotor area (48.78 m²) - D is the diameter of the rotor (which we want to find) Rearrange the formula to solve for D: D = 2 × √(A ÷ π) Next, plug in the given values and calculate D: D ≈ 2 × √(48.78 ÷ 3.1416) D ≈ 7.87 m
05

Compare diameter to familiar objects

The diameter of the ocean current rotor is found to be approximately 7.87 meters. To give context, this is roughly the length of a school bus or around half the width of a basketball court. This comparison can help us visualize the scale of the ocean current rotor.

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Most popular questions from this chapter

If it were possible to achieve \(60 \%\) of the theoretical maximum heat engine efficiency for a geothermal plant, how deep would it have to drill to access high enough temperatures to match the \(35 \%\) efficiency of fossil fuel power plants if the thermal gradient is \(25^{\circ} \mathrm{C} / \mathrm{km} ?\) Have we drilled this far before?

On the basis of fluid power scaling as the cube of velocity, show the supporting math for the claim in the text that a water current at \(1 \mathrm{~m} / \mathrm{s}\) delivers the equivalent power (per rotor area) as a wind speed of about \(9 \mathrm{~m} / \mathrm{s}\).

Let's imagine waves hitting the entire \(2,000 \mathrm{~km}\) Pacific coastline of the U.S. that are different from those evaluated in the text. This time, the waves have \(50 \mathrm{~m}\) wavelength, arriving every 10 seconds, and \(2 \mathrm{~m}\) crest-to-trough amplitude. How much power does the coast receive under these more active conditions?

A typical college campus is probably about one square kilometer in area. How much power out of Earth's 44 TW geothermal budget passes through the campus area, assuming uniform distribution across the \(5.1 \times 10^{8}\) square kilometers of earth's surface? How does this compare to a typical college electrical demand of about \(20 \mathrm{MW} ?\)

What are your thoughts on whether we should "mine" geothermal heat in a way that could theoretically last hundreds or thousands of years, but not tens of thousands of years before depleting the resource? Such activity would not be strictly sustainable in the long haul, but could seem abundant for many generations. Should we care about this? Where do you land?
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